Destroying Array

                                                                                                              C -Destroying Array

                          Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

Description

You are given an array consisting of n non-negative integersa₁, a₂, ..., a_n.

You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from1 ton defining the order elements of the array are destroyed.

After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be0.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.

The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹).

The third line contains a permutation of integers from 1 to n — the order used to destroy elements.

Output

Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after firsti operations are performed.

Sample Input

Input

41 3 2 53 4 1 2

Output

5430

Input

51 2 3 4 54 2 3 5 1

Output

65510

Input

85 5 4 4 6 6 5 55 2 8 7 1 3 4 6

Output

18161188660

Sample Output

Hint

Consider the first sample:

1. Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum5 consists of one integer5.
2. Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum4 consists of two integers1 3.
3. First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum3 consists of one integer3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to0.

题意：

起先你有n个正整数a1,a2,...,an，现在你要将这个数组里的数一个接一个的破坏掉，直到最后每个数都被
破坏，对于每次破坏，求出当前最大连续段和，要求这些段不能包含被破坏的数

题解：

正着删除不好处理，就反过来求了，相当于从一个空序列往上面添加数值，加一次，求一次线段的最大和，使用离线处理，每次与当前值和前一个答案来更新答案！

加入某个值后，使用用并查集进行处理，令a[i]是 i 这个点的权值，那么先判断i+1的位置，如果有值的话，就把i+1这一系列线段与i位置上的值合并，再判断i-1位置。

#include<stdio.h>#include<iostream>#include<string.h>using namespace std;long long int a[100010],b[100010],vis[100010];long long int ba[100010],ba1[100100],ans[100010];int zhao(int i){    while(ba[i]!=i)    {        i=ba[i];    }    return i;}int main(){    int n;    while(scanf("%d",&n)!=-1)    {        for(int i=1;i<=n;i++)            scanf("%lld",&a[i]);        for(int i=1;i<=n;i++)            scanf("%lld",&b[i]);        for(int i=1;i<=n;i++)            ba[i]=i;            memset(vis,0,sizeof(vis));           ans[n]=0;        for(int i=n;i>1;i--)        {            ba1[b[i]]=a[b[i]];            if(vis[b[i]-1]==1)            {                int f1,f2;                f1=zhao(b[i]);                f2=zhao(b[i]-1);                if(f1!=f2)                {                ba[f1]=f2;                ba1[f2]=ba1[f2]+ba1[f1];                }            }            if(vis[b[i]+1]==1)            {                int f1,f2;                f1=zhao(b[i]);                f2=zhao(b[i]+1);                if(f1!=f2)                {                ba[f1]=f2;                ba1[f2]=ba1[f2]+ba1[f1];                }            }            vis[b[i]]=1;            ans[i-1]=max(ans[i],ba1[zhao(b[i])]);        }        for(int i=1;i<=n;i++)            printf("%lld\n",ans[i]);    }}